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Tutorial set-4
Elementary University Physics II (Carleton University)
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Tutorial set-4
Elementary University Physics II (Carleton University)
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PHYS 1008 2017W
Tutorial #4
All sections
1.
A glass block with index of refraction
n
1
= 2
.
00
is submerged in an unknown
liquid with index of refraction
n
2
.
(a)
A light ray inside the glass block is incident on the glass-liquid interface
at an angle of
θ
= 40
.
0
◦
as shown in figure (a).
If the ray undergoes
total internal reflection, what is the maximum value of
n
2
?
(b)
A second light ray inside the glass block is incident on the glass-liquid
interface at an angle of
θ
= 35
.
0
◦
as shown in figure (b).
In this case,
the transmission angle is
90
.
0
◦
so that a ray is in the interface plane
between the glass and the liquid. What is the critical angle and what
is the index of refraction
n
2
?
(c)
A third light ray inside the glass block is incident on the glass-liquid
interface at an angle of
θ
= 30
.
0
◦
as shown in figure (c).
What is the
transmission angle
θ
?
Solution:
(a) First note that the critical angle is necessarily less than or equal to
40
.
0
◦
, that is
θ
c
≤
40
.
0
◦
. If we saturate this bound, we can find the maximum value
for
n
2
.
n
2
max
=
n
1
sin 40
.
0
◦
= (2
.
00) sin 40
.
0
◦
= 1
.
29
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Thus
n
2
<
1
.
29.
(b) Since the transmission angle is 90
.
0
◦
, we simply have
θ
c
= 35
.
0
◦
.
The index of
refraction
n
2
is then
n
2
=
n
1
sin 35
.
0
◦
= 2
.
00 sin 35
.
0
◦
= 1
.
15
(c) Using Snell’s law, we get
n
1
sin 30
.
0
◦
=
n
2
sin
θ
θ
= arcsin
parenleftbigg
n
1
sin 30
.
0
◦
n
2
parenrightbigg
= arcsin
parenleftbigg
(2
.
00) sin 30
.
0
◦
1
.
15
parenrightbigg
= 60
.
7
◦
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2.
Two converging lenses are placed 60.0 cm apart. An object is placed 80.0
cm to the left of the first lens. The focal length of the first lens is
f
1
= 20
.
0 cm
and the focal length of the second lens is
f
2
= 15
.
0 cm
.
(a)
Where is the image from the first lens formed?
(b)
What is the magnification of the first lens?
(c)
Where is the image from the second lens formed?
(d)
What is the magnification of the second lens?
(e)
What is the total magnification?
(f)
Is the final image upright or inverted?
(g)
Draw a ray diagram of this situation.
Solution:
(a) Use the thin-lens equation for the first lens
1
p
1
+
1
q
1
=
1
f
1
q
1
=
bracketleftbigg
1
f
1
-
1
p
1
bracketrightbigg
−
1
q
1
=
bracketleftbigg
1
20
.
0
-
1
80
.
0
bracketrightbigg
−
1
q
1
= 26
.
7 cm
i.e. 26.7 cm to the right of the first lens.
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(b) The magnification of the first lens is
m
1
=
-
q
1
p
1
=
-
26
.
7
80
.
0
=
-
0
.
333
(c) First we note that
p
2
= 60
.
0
-
q
1
= 60
.
0
-
26
.
7 = 33
.
3 cm. Then
1
p
2
+
1
q
2
=
1
f
2
q
2
=
bracketleftbigg
1
f
2
-
1
p
2
bracketrightbigg
−
1
q
2
=
bracketleftbigg
1
15
.
0
-
1
33
.
3
bracketrightbigg
−
1
q
2
= 27
.
3 cm
i.e. 27.3 cm to the right of the second lens.
(d) The magnification of the second lens is
m
2
=
-
q
2
p
2
=
-
27
.
3
33
.
3
=
-
0
.
819
(e) The total magnification is
m
t
=
m
1
m
2
= (
-
0
.
333)(
-
0
.
819)
= 0
.
273
(f) Since the total magnification is positive, the final image is upright.
(g) There are many different possible ray diagrams for this situation. Below we show
4 different examples.
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Normal line
Li=
P
S
Z
Ls
Normal line
B
Le = 30°
● - Dot represents a pin
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Step 9
C
Step 10
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